Problem 14 - Master Resource Book in JEE Main Physics

Question

A bomb is dropped on an enemy post by an aeroplane flying horizontally with a velocity of \(60 \mathrm{kmh}^{-1}\) and at a height of \(490 \mathrm{~m}\). At the time of dropping the bomb, how far the aeroplane should be from the enemy post so that the bomb may directly hit the target? (a) \(\frac{400}{3} \mathrm{~m}\) (b) \(\frac{500}{3} \mathrm{~m}\) (c) \(\frac{1700}{3} \mathrm{~m}\) (d) \(498 \mathrm{~m}\)

Step-by-Step Solution

Answer

Option (b) \(\frac{500}{3} \text{ m}\).

Step 1: Convert velocity to consistent units

The velocity is given as \(60 \text{ km/h}\). First, convert this velocity to \(\text{m/s}\) because the other units in the problem are in meters and seconds. \[1 \text{ km/h} = \frac{1000}{3600} \text{ m/s} = \frac{5}{18} \text{ m/s}\]Thus, \[60 \text{ km/h} = 60 \times \frac{5}{18} \text{ m/s} = \frac{50}{3} \text{ m/s}\]

Step 2: Calculate time of flight

The bomb is dropped from a height \( h = 490 \text{ m} \). Using the formula for the time of flight of a free-falling object, \[ t = \sqrt{\frac{2h}{g}}\]where \( g = 9.8 \text{ m/s}^2 \). Plug in the values:\[ t = \sqrt{\frac{2 \times 490}{9.8}} = \sqrt{100} = 10 \text{ seconds}\]

Step 3: Calculate horizontal distance

Now, use the horizontal velocity and the time calculated to find the horizontal distance \( s \) that the aeroplane should be from the enemy post. The horizontal distance is given by the formula:\[ s = v \times t\]where \( v = \frac{50}{3} \text{ m/s} \) and \( t = 10 \text{ seconds} \).So,\[ s = \frac{50}{3} \times 10 = \frac{500}{3} \text{ m}\]

Step 4: Match with given options

The computed horizontal distance is \(\frac{500}{3} \text{ m}\). Now, look at the provided options:(a) \(\frac{400}{3} \text{ m}\)(b) \(\frac{500}{3} \text{ m}\)(c) \(\frac{1700}{3} \text{ m}\)(d) \(498 \text{ m}\)The correct match is option (b): \(\frac{500}{3} \text{ m}\).

Key Concepts

Horizontal and Vertical Motion

In projectile motion, we often deal with movements in two dimensions simultaneously. Generally, this involves the horizontal and vertical components of motion. Each component acts independently of the other and they are influenced by different factors.

• Horizontal Motion: This component is characterized by uniform motion because there are no external forces like gravity acting parallel to the horizontal axis. In our exercise, the aeroplane flies horizontally at a constant speed. This means the horizontal velocity remains steady throughout the flight.


• Vertical Motion: On the other hand, vertical motion is affected by gravity. This results in accelerated motion. As the bomb drops, it speeds up due to the gravitational pull. The vertical component involves calculating the time it takes for the bomb to reach the ground from the given height.

Understanding these concepts separately helps us tackle problems involving projectile motion more effectively, allowing us to analyze each component individually before combining them for the final result.

Time of Flight Calculation

Calculating the time of flight is essential for determining how long an object remains in projectile motion, particularly the bomb in the exercise. This tells us how long it takes for the bomb to descend from the airplane to the target ground.
The formula used here is derived from the equations of motion under gravity. We use:\[t = \sqrt{\frac{2h}{g}} \]where:

• \( h \) is the height from which the bomb is dropped


• \( g \) is the acceleration due to gravity, approximately \(9.8 \text{ m/s}^2\).

Inserting the values from the exercise:

• Given height, \( h = 490 \text{ m} \)


• \( t = \sqrt{\frac{2 \times 490}{9.8}} = 10 \text{ seconds} \)

Understanding this formula is critical because the time of flight connects the vertical and horizontal motions, allowing us to measure the duration the bomb is in the air before hitting the target.

Velocity Conversion

Velocity conversion is a fundamental skill when working with different units in problem-solving. In physics, consistency in units is crucial to avoid errors.
In this exercise, the initial velocity is given in kilometers per hour (km/h), but all other quantities use meters and seconds. We convert the horizontal velocity to meters per second (m/s) to maintain uniformity.
To convert, you apply the conversion factor:\[1 \text{ km/h} = \frac{1000}{3600} \text{ m/s} = \frac{5}{18} \text{ m/s}\]Thus, for \(60 \text{ km/h}\):

• \( 60 \times \frac{5}{18} = \frac{50}{3} \text{ m/s} \)

This conversion enables a seamless integration of values in formulas for further calculations, ensuring that all measurements remain compatible throughout the projectile motion analysis.

Distance Calculation

Calculating the horizontal distance ensures the bomb accurately hits the target on the ground once released. This calculation involves the constant horizontal velocity and the calculated time of flight.
The formula to find horizontal distance \( s \) is:\[s = v \times t\]where:

• \( v \) is the horizontal velocity in meters per second (converted from km/h), \( \frac{50}{3} \text{ m/s} \)


• \( t \) is the time of flight, calculated as \(10\) seconds.

In the exercise, substituting these values gives:

• \( s = \frac{50}{3} \times 10 = \frac{500}{3} \text{ m} \)

This distance represents how far the airplane must be from the target when releasing the bomb, ensuring it impacts the enemy post directly. Understanding and applying this distance calculation is vital to solving projectile motion problems and achieving precise targeting.